Answer:
while(is_sunny=="n")
Explanation:
The loop keeps repeating itself until a certain condition is met.
Here while loop keeps executing until value of is_sunny is not equal to 'n'
When the value of is_sunny is not equal to n then the loop stops.
So lets explain this logic with a chunk of code:
#include <iostream> //to use input output functions
using namespace std; //to identify objects like cin cout
int main() { //start of main function
string is_sunny = "n"; //value of is_sunny is set to n
cout<<"Enter value of is_sunny: "; // prompts user to enter value of is_sunny
cin>>is_sunny; // reads value of is_sunny from user
while(is_sunny=="n") // keeps iterating until value of is_sunny is not equal to n
{ cout<<"keep executing until is_sunny is not equal to n"<<endl;//display this message with each iteration
cout<<"Enter value of is_sunny: "; //keeps prompting to enter is_sunny
cin>>is_sunny; } } //keeps reading value of is_sunny from user
Now if the user keeps entering "n" as value of is_sunny then the loop keeps repeating. When the user enters any string other than n then the loop breaks. The output of the program is attached.
You click the preorder option, normally on apps or books there is a button that says "pre-order"
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Answer:
Explanation:
a) First, we need to determine the pipeline stage amounting to the maximum time. In the given case, the maximum time required is 2ns for MEM. In addition, the pipeline register delay=0.1 ns.
Clock cycled time of the pipelined machine= max time+delay
=2ns+0.1 ns
=2.1 ns
b) For any processor, ideal CPI=1. However, since there is a stall after every four instructions, the effective CPI of the new machine is specified by:
1+(1/4)=1.25
c) The speedup of pipelined machine over the single-cycle machine=avg time per instruction of single cycle/avg time per instruction of pipelined.
Single cycle processor:
CPI=1
Clock period=7 ns
Pipelined processor:
Clock period=2.1 ns
CPI=1.25
Therefore, speedup=7*1/(2.1*1.25)=7/2.625
= 2.67
d) As the number of stages approach infinity, the speedup=k where k is the number of stages in the machine
Answer:
Radial gradients radiate from a center focal point. Both can be edited for color, alpha, and position within a fill or stroke. You can add up to 16 colors to a gradient, precisely control the location of the gradient focal point, and apply other parameters. A gradient behaves like any other fill or stroke.
Answer:
B. 0
Explanation:
The computation happens in the following sequence:
1) The operands 4.5 and 6.4 are added : 4.5+6.4 = 10.9
2) When this value is cast to int datatype, it becomes 10
3) 3/10 = 0.3
4) x in a variable of type int. So when a value of 0.3 is assigned to x, it is stored as 0.
If we now display the value of x using printf or cout statement, we should expect to see 0 as the value printed on the console.